Subfield Criteria - Proof or Counterexample I am interested in whether the following claim is true for all fields $F$:
Conjecture:
A subset $X\subset F$ is a subfield if and only if (1) $1\in X$, (2) $x,y\in X\Rightarrow x-y\in X$; and (3) $x\in X\setminus\{0\}\Rightarrow x^{-1}\in X$.
It isn't too hard to prove true if char$F\neq2$, and I can also show it holds if $F$ is finite. The case of an infinite field of characteristic $2$ remains elusive, though. I can neither prove it nor find a counterexample. Can anyone else give it a go?
Some notes: conditions (1), (2) and (3) are certainly independent, hence my interest in the problem. The following are consequences of the conditions:


*

*If $x,y\in X$ then $(xy)^2,xy^2\in X$

*If $x\in X$ then $x^n\in X$ for all $n\in\mathbb{Z}$


In case you were after the work so far:
Clearly conditions (1) and (2) imply $X$ is a subgroup. Suppose $x\in X$. If $x\in\{0,1\}$ then $x^2=x\in X$. If not, $x^{-1},(1-x)^{-1}\in X$, so $x^2=x-[x^{-1}+(1-x)^{-1}]^{-1}\in X$. Hence $X$ is closed under squares.
Assume char$F\neq2$. Then if $x,y\in X$, $2xy=(x+y)^2-x^2-y^2\in X$, so either $xy=0\in X$ or $(2xy)^{-1}\in X$, in which case $xy=[(2xy)^{-1}+(2xy)^{-1}]^{-1}\in X$. This proves the conjecture if char$F\neq2$.
Assume char$F=2$. Then if $x,y,z\in X$ with $x\notin\{y^{-1},z^{-1}\}$ and $y,z,0$ all distinct, we have $\omega:=(x+y^{-1})^{-1}+(x+z^{-1})^{-1}\in X$. Simplifying gives $\omega=[y(1+xz)+z(1+xy)][(1+xy)(1+xz)]^{-1}=(y+z)[1+x(y+z)+x^2yz]^{-1}$, and so
$$f(x,y,z):=x^2yz(y+z)^{-1}=\omega^{-1}-x-(y+z)^{-1}\in X$$
for all $x,y,z\in X$. Thus if $x,y\in X$, we have
$$x^2y(y+1)=f(x,y,y+1)\in X,\quad y(x^2+y)=f(x,y,x^2+y)\in X\Rightarrow x^2y\in X\Rightarrow x^2y^2\in X.$$
If $|F|<\infty$, then $|F|=2^n$ for some $n$ and $F\setminus\{0\}$ is a multiplicative group of order $2^n-1$. Since $X$ is closed under squares, induction gives us
$$xy=(xy)^{2^n}\in X.$$
If $|F|=\infty$, I'm stuck. Since $x\in X$, $x^3=x^2x\in X$, and if $x^{n-2}\in X$ then $x^n=x^{n-2}x^2\in X$. Hence by induction $x^n\in X$ for all $n\in\mathbb{N}$ (since $X$ is closed under squares and cubes), and so by taking inverses $x^n\in X$ for all $n\in\mathbb{Z}$.
 A: Let $$A=\biggl\{\sum a_{n,m}x^ny^m\in \mathbb F_2[x,y]\biggm| a_{n,m}= 0\text{ if $mn$ is odd}\,\biggr\}\subset \mathbb F_2[x,y]$$ be the additive subgroup of $\mathbb F_2[x,y]$ that is generated by all monomials with at least one of the two exponents even.
Then one sees immediately that
$$\tag1 1,x,y\in A.$$
In characteristic $2$, we have $(a+b)^2=a^2+b^2$, hence 
if $g\in\mathbb F_2[x,y]$, then all monomials $x^my^n$ occuring in $g^2$ have  both exponents even. From this,
$$\tag2f\in A, g\in \mathbb F_2[x,y]\implies fg^2\in A$$
and
$$\tag30\ne g\in \mathbb F_2[x,y]\implies xyg^2\notin A.$$
Now let $$ X=\bigl\{\,q\in\mathbb F_2(x,y)\bigm | g^2q\in A\text{ for some nonzero $g\in \mathbb F_2[x,y]$}\,\bigr\}\subset\mathbb F_2(x,y).$$
Then 


*

*Because of $(1)$, we have $1\in X$.

*Because $A$ is a group and because of $(2)$, we have that $q_1,q_2\in X$ implies $q_1-q_2\in X$. Indeed, if $q_1g_1^2,q_2g_2^2\in A$ with $g_1,g_2\ne 0$, then $(q_1-q_2)(g_1g_2)^2=q_1g_1^2\cdot g_2^2-q_2g_2^2\cdot g_1^2\in A$.

*Because of $(2)$, $0\ne q\in X$ implies $q^{-1}\in X$. Indeed, $qg^2\in A$ with $g\ne 0$ implies $q^{-1}\cdot (qg^2)^2=qg^2\cdot g^2\in A$.


This shows that $X$ has the properties described in the problem statement.
From $(3)$ we have $xy\notin X$, whereas $(1)$ shows $x,y\in X$. Hence $X$ is not a field.
